Question

An archer shoots an arrow at a 74.0 m distant target, the bull's-eye of which is at same height as the release height of the arrow.
(a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 33.0 m/s? (Although neglected here, the atmosphere provides significant lift to real arrows.)
°
(b) There is a large tree halfway between the archer and the target with an overhanging branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?
under
over

Answer 1

A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °

B. The arrow will go over the branch.

A. How to determine the angle

• Range (R) = 74 m
• Initial velocity (u) = 33 m/s
• Acceleration due to gravity (g) = 9.8 m/s²
• Angle (θ) = ?

R = u²Sine(2θ) / g

74 = 33² × Sine (2θ) / 9.8

Cross multiply

74 × 9.8 = 33² × Sine (2θ)

725.2 = 1098 × Sine (2θ)

Divide both sides by 1098

Sine (2θ) = 725.2 / 1098

Sine (2θ) = 0.6605

Take the inverse of sine

2θ = Sine⁻¹ 0.6605

2θ = 41.3

Divide both sides by 2

θ = 41.3 / 2

θ = 20.7 °

B. How to determine if the arrow will go over or under the branch

To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:

• Initial velocity (u) = 33 m/s
• Acceleration due to gravity (g) = 9.8 m/s²
• Angle (θ) = 20.7 °
• Maximum height (H) = ?

H = u²Sine²θ / 2g

H = [33² × (Sine 20.7)²] / (2 ×9.8)

H = 6.94 m

Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).

Therefore, we can conclude that the arrow will go over the branch

Learn more about projectile motion:

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